Question

Air contains 23% oxygen and 77% nitrogen by weight. The percentage of oxygen by volume is:
a.) 28.1
b.) 20.7
c.) 21.8
d.) 23.0

Answer 1

Hint: We can consider any value of the mass of air we have and proceed accordingly. The percentage of oxygen by volume will be equal to the volume of oxygen divided by the total volume of air.

Complete step by step answer: For our convenience, let the mass of air we have be equal to 100gm.
Therefore, the mass of oxygen in 100 grams of air = 23gm
And the mass of nitrogen in 100 grams of air = 77gm

Now, we know that,
The molecular mass of \[{O_2}\] = 32 grams
And the molecular mass of \[{N_2}\] = 28 grams.
Therefore, number of moles of \[{O_2}\] = \[\dfrac{{23}}{{32}}\]
And the number of moles of \[{N_2}\] = \[\dfrac{{77}}{{28}}\]

As one mole of a gas occupies 22.4L volume at Standard Temperature and Pressure conditions (STP)
So,
Volume occupied by moles of oxygen = \[\dfrac{{23}}{{32}}\] $\times$ 22.4L = 16.1L

Volume occupied by moles of nitrogen = \[\dfrac{{77}}{{28}}\] $\times$ 22.4L= 61.6L

Therefore, total volume of the air = 16.1L + 61.6L = 77.7L

So, percentage by volume of Oxygen = \[\dfrac{{16.1}}{{77.7}} \times 100\% \] = 20.7%

Hence, the correct answer is (B) 20.7%.

Note: Remember that the percentage by volume of a substance is the ratio of volume by volume, not the ratio of mass by volume.